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Continuous Probability
Distributions
•Continuous Random Variable:
Values from Interval of Numbers
Absence of Gaps
•Continuous Probability Distribution:
Distribution of a Continuous Variable
•Most Important Continuous Probability
Distribution: the Normal Distribution
The Normal Distribution
• ‘Bell Shaped’
• Symmetrical
f(X)
• Mean, Median and
•
Mode are Equal
•Random Variable has
•
Infinite Range
m
Mean
Median
Mode
X
The Mathematical Model
2
f(X) =
1
2p s
e
(-1/2)((X- m)/s)
f(X) =
frequency of random variable X
p
=
3.14159;
s
=
population standard deviation
X
=
value of random variable (- < X < )
m
=
population mean
e = 2.71828
Many Normal Distributions
There are
an Infinite
Number
Varying the Parameters s and m, we obtain
Different Normal Distributions.
Which Table?
Each distribution
has its own table?
Infinitely Many Normal Distributions Means
Infinitely Many Tables to Look Up!
The Standardized Normal
Distribution
Standardized Normal Probability
Table (Portion)
m Z = 0 and s Z = 1
Z
.00
.01
.0478
.02
0.0 .0000 .0040 .0080
0.1 .0398 .0438 .0478
0.2 .0793 .0832 .0871
Z = 0.12
0.3 .0179 .0217 .0255
Probabilities
Shaded Area
Exaggerated
Standardizing Example
Normal
Distribution
6
.
2

5
X

m
Z

 0.12
s
10
Standardized
Normal Distribution
s = 10
sZ = 1
m = 5 6.2 X
m = 0 .12
Shaded Area Exaggerated
Z
Example:
P(2.9 < X < 7.1) = .1664
z 
Normal
Distribution
z 
xm
s
x  m
s

2 .9  5
  . 21
10

7 .1  5
 . 21
10
Standardized
Normal Distribution
s = 10
s =1
.1664
.0832 .0832
2.9 5 7.1 X
-.21 0 .21
Shaded Area Exaggerated
Z
Example: P(X  8) = .3821
z
xm
s
Normal
Distribution
85

 .30
10
Standardized
Normal Distribution
s = 10
s =1
.5000
.1179
m =5
8
X
.3821
m = 0 .30 Z
Shaded Area Exaggerated
Finding Z Values
for Known Probabilities
What Is Z Given
P(Z) = 0.1217?
.1217
s =1
Standardized Normal
Probability Table (Portion)
Z
.00
.01
0.2
0.0 .0000 .0040 .0080
0.1 .0398 .0438 .0478
m = 0 .31 Z
Shaded Area
Exaggerated
0.2 .0793 .0832 .0871
0.3 .1179 .1217 .1255
Finding X Values
for Known Probabilities
Normal Distribution
Standardized Normal Distribution
s = 10
s =1
.1217
m =5
?
X
.1217
m = 0 .31
X  m + Zs = 5 + (0.31)(10) = 8.1
Shaded Area Exaggerated
Z
Assessing Normality
• Compare Data
Characteristics
• to Properties of Normal
• Distribution
• Put Data into Ordered Array
• Find Corresponding Standard
•
Normal Quantile Values
• Plot Pairs of Points
• Assess by Line Shape
Normal Probability Plot
for Normal Distribution
90
X 60
Z
30
-2 -1 0 1 2
Look for Straight Line!
Normal Probability Plots
Left-Skewed
Right-Skewed
90
90
X 60
X 60
Z
30
-2 -1 0 1 2
-2 -1 0 1 2
Rectangular
U-Shaped
90
90
X 60
X 60
Z
30
-2 -1 0 1 2
Z
30
Z
30
-2 -1 0 1 2
Estimation
•Sample Statistic Estimates Population Parameter
_
•
e.g. X = 50 estimates Population Mean, m
•Problems: Many samples provide many estimates of the
Population Parameter.
• Determining adequate sample size: large sample give better
estimates. Large samples more costly.
•
How good is the estimate?
•Approach to Solution: Theoretical Basis is Sampling
Distribution.
Properties of Summary
Measures
• Population Mean Equal to
• Sampling Mean m x  m
• The Standard Error (standard deviation)
of the Sampling distribution is Less than
Population Standard Deviation
• Formula (sampling with replacement):
s x_ =
s
n
As n increase,
s x_
decrease.
Central Limit Theorem
As Sample
Size Gets
Large
Enough
Sampling
Distribution
Becomes
Almost Normal
regardless of
shape of
population
X
X
Population Proportions
• Categorical variable (e.g., gender)
• % population having a characteristic
• If two outcomes, binomial distribution
– Possess or don’t possess characteristic
• Sample proportion (ps)
X
number of successes
Ps 

n
sample size
Sampling Distribution of
Proportion
• Approximated by
normal distribution
Sampling Distribution
– n·p  5
P(ps)
 – n·(1 - p)  5
.3
.2
• Mean m P  p
• Standard error
sP 
p  1  p )
n
.1
0
0
.2
.4
.6
8
1
p = population proportion
ps
Standardizing Sampling
Distribution of Proportion
Z @
ps - m p
sp
Sampling
Distribution
=
ps - p
p( 1  p )
n
Standardized
Normal Distribution
sp
s=1
mp
ps
m =0
Z